circular motion class 11 physics

1. Circular motion
Angular displacement
The angular displacement of the object moving in a circular direction is defined as the
angle traced out by the radius vector at the center of the circular path in a given time. The
SI-unit of angular displacement is radian.
Angular velocity
The time rate of angular displacement is called angular velocity and is denoted by ω.
Mathematically, ωav = angular displacement / time interval
= (θ2 – θ1)/(t2 – t1)
= ∆θ/∆t

2. • Time period
• The period of an object in circular motion is defined as the time taken to
complete one revolution. It is denoted by T. since in time T, the object
completes one revolution i.e. an angle of 2π radians, angular velocity of an
object is
• ω = 2π/T
•
• Frequency
• The number of revolutions completed per second by an object in circular
motion is called frequency and is denoted by f. the relation between the
frequency, f and period, T is
• F = 1/T
• Therefore, we have
• The unit of frequency is hertz, Hz.

3. •
• Angular acceleration
• The rate of change of angular velocity with respect to time is called angular
acceleration. It is denoted by α.
• .: Angular acceleration, α = Change in angular velocity/time taken
•
• If ω0 and ω are the initial and final angular velocities and t is the time taken to
change the angular velocity, then,
• α = (ω – ω0)/ t
• ω = ω0 + α. T
•
• Angular acceleration is measured in radian per second square (rad. s-2) and its
dimensional formula is [M0L0T-2].
• Angular acceleration is also given by, dω/dt

4. Relation between Linear Velocity and Angular Velocity
• Suppose a particle of mass m moving in a circular path of radius r with constant speed v. let θ be the
angular displacement when particle goes from points P to Q in time t as shown in the figure. If s is
the arc of length PQ, then
• θ = s/r
• or, s = r. θ
•
• Differentiating both sides with respect to time, we have
• ds/dt = r. dθ/dt

5. • As r is constant, v = ds/dt, and angular velocity, ω = dθ/dt
• Then, above equation can be written as
• V = r. ω
• This is the relation between linear velocity and angular velocity.
•
• Since linear acceleration is given by a = dv/dt
• And angular acceleration is given by α = dω/dt
• Differentiating both sides w.r.t time in V = r. ω, we get
• a = r. α

6. Expression for Centripetal Acceleration
• Consider a particle moving with a constant speed v in a circular path
of radius r with center O. At a point A on its path, its velocity vA is
along the tangent at A and at B, its velocity vB is along the tangent at
B. since the direction of the velocity at A and B are different, an
acceleration occurs from A to B.
• Let ∆t be the time taken to move from A to B covering a small
displacement Δl and angular displacement Δθ.

7. • As shown in figure vA and vB are redrawn where Δv represents the
change in velocity, Δv = vB - vA, which form three sides of a triangle
PQR. Two triangles OAB and PQR are similar since both are isosceles
triangles and angles labelled Δθ are the same. So, the ratios of the
corresponding sides of such triangles are equal.

8. • Thus,
• Δv/v = Δl/r
• or, Δv = Δl/r . Δl
•
• The rate of change of velocity is
• Δv/Δt = v/t . Δl/Δt
•
• For Δt->0,
• Limit Δt-> 0 Δv/Δt = a and,
•
• Limit Δt-> 0 Δl/Δt = v
• Then, the above equation is written as
• a = v/r . v
• or, a = v2/ r
•
• Since v = r. ω, the centripetal acceleration can be written as
• a = r. ω2

9. Centripetal Force
Centripetal force is the force required to move a
body uniformly in a circle. This force acts along
the radius and is directed towards the center of
the circle

10. • When a body moves in a circle, its direction of motion at any instant is along
the tangent to the circle at that instant. As shown in figure, the direction of
velocity of the body goes in changing continuously and due to this change,
there is an acceleration called the centripetal acceleration given as,
• a = v2/ r = r. ω2
• Where v is linear velocity, ω is angular velocity of the body, and r is radius of
the circular path.
•
• As F = m. a, centripetal force is
• F = mass * acceleration
• = mv2/r = m. r. ω2
• And this force acts along the radius and is directed towards the center of the
circle.
•

11. Application of Centripetal force
• 1. Motion of a car on a Level Curved Path
• Let F1 and F2 be the forces of friction between the wheels and the road,
directed towards the center of the horizontal curved track of radius r as
shown in the figure. From the law of Friction,
• F1 + F2 = µ (R1 + R2) …... (i)
• Where µ is the coefficient of sliding friction between tires and the road
and R1 and R2 are the forces of normal reaction of the road on the
wheels.

12. As there is no motion of the car in the vertical direction, we have
R1 + R2 = m. g ….... (ii)
As the force of friction provides the necessarily centripetal force, then
F1 + F2 = m. v2/ r …...... (iii)
So, From equation (ii) and (iii), we have
µmg = mv2/ r
or, µ = v2/ rg
.: v = √µ rg ….... (iv)
This equation gives the maximum
velocity with which the car can take a
circular turn of radius r and coefficient of
friction µ. This shows that much more
friction is required as the turning speed
is to be increased and if the velocity is
greater than determined by equation (iv),
the skidding of the car occurs and moves
out of the road. This equation holds for a
vehicle of any mass from a bicycle to a
heavy truck.

13. • 2. Banking of road
• We can not always depend on the friction to move a car or other vehicles
around a curve, especially if the road is icy or wet. In order not to depend on
friction or to reduce wear on the tires on road, the roads are banked as shown
in the figure.

14. • The effect of banking is to tilt the normal toward the center of the curved road, so
that it's inward horizontal component can supply the required centripetal force. So,
resolving (R1 + R2) into two components (R1 + R2)sinθ towards the center and (R1 +
R2)cosθ where θ is the banking angle or angle of inclination of the plane with the
horizontal, we have
• (R1 + R2) sinθ = mv2/r
•
• And vertically
• (R1 + R2) cosθ = mg
•
• From these two equations, we have
• (R1 + R2) sinθ/(R1 + R2) cosθ = mv2/ r * 1/ mg
• or, Tanθ = v2/ rg
•
• The above equation gives the angle for banking for a speed v and a curved track of
horizontal radius r. so the tangent of the banking angle is directly proportional to
the sped and inversely proportional to the radius of the path.

15. • 3. Bending of a Cyclist
• When a cyclist takes a turn, he bends toward the center of the circular path
and the horizontal component of normal reaction directed towards the center
provides the necessary centripetal force for the motion. Suppose a cyclist of
mass m going in a circular path of radius r with constant speed v tilting at an
angle of θ with the vertical. Then R can be resolved into two rectangular
components: R cosθ in the vertically upward direction and R sinθ, along the
horizontal, towards the center of the circular track as shown in the figure. In
equilibrium, R cosθ balances the weight mg of the cyclist and R sinθ provides
the necessary centripetal force, mv2/ r.

17. • So,
• R cosθ = mg …..... (i)
•
• And,
• R sinθ = mv2/ r …...... (ii)
•
• Dividing equation (ii) by equation (i), we get
• Tanθ = v2/ rg
• This relation gives the angle through which a cyclist tilts with the vertical
while going in a circular road of radius r at a speed of v.
•